# Coin Tossing Game Optimal Strategy Part 2

Coin Tossing Game Optimal Strategy

Here I'd like to complicate the question slightly. This part stays the same:

You start off with £100 and you toss a coin 100 times. Before each toss you choose a stake S which cannot be more than your current balance x (so your maximum stake for the first toss is £100). If the coin comes up heads, you win 2S and your new balance is x+2S. If it comes up tails, you lose your stake and have x−S.

This time, though. Instead of maximizing the expected profit $E[P]$, we want to maxmise $E[\log(100+P)]$. In both cases the initial balance of £100 is not included in any profit. How should we choose our bets in this case? Now, my first thought and indeed my answer was that whichever strategy maximizes the first case must also maximize the second.. but the very fact that it was asked makes me doubt myself? Any thoughts greatly appreciated!

Boris

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The two questions are quite different. In this case, you cannot ever afford to bet all your money, as the log goes to $-\infty$ –  Ross Millikan Nov 28 '12 at 16:31

On the last toss, you want to maximize

$$\frac12\left(\log(x+2S)+\log(x-S)\right)\;,$$

where $x$ is your balance before the last toss. Differentiating with respect to $S$, setting to zero and solving for $S$ yields $S=x/4$. The expected value for this optimal bet is

$$\frac12\left(\log(x+x/2)+\log(x-x/4)\right)=\log x+\log3-\frac32\log 2\approx\log x+0.06\;.$$

Thus the expected value is simply the value $\log x$ you had before plus about $0.06$. It follows that on the previous toss you should also maximize the expected value of the logarithm of the balance, and likewise for all previous tosses, so on each toss you should bet one quarter of your current balance, and the expected value after $100$ tosses is

$$\log100+100\left(\log3-\frac32\log 2\right)\approx10.49\;.$$

(To compare with Ross' answer: I'm using natural logarithms whereas he uses decimal logarithms.)

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For one toss, you should bet $25$. We want to maximize $\frac 12 \log (100+2S) + \frac 12 \log (100-S)$. Taking the derivative and setting to zero shows the maximim comes at $S=25$. Using base $10$ logs, this gives an expectation of about $2.026$ compared to the $2$ we started with.

For a two toss game, this says the last bet should be one quarter of the current bankroll. Letting $S$ be the bet on the first turn. After one turn we have either $100+2S$ or $100-S$.Betting one quarter of whatever w have, we want to maximize $\log (150+3S) + \log (75+\frac {3S}2)+\log (150-\frac {3S}2)+\log (75-\frac {3S}4)$ which Alpha says comes at $25$. This gives an expectation of about $2.051$. I suspect this means you should always bet one quarter of the bankroll, but haven't proved it.

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